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Ammonia and potassium iodide solutions are added to an aqueous solution of \(\mathrm{Cr}\left(\mathrm{NO}_{3}\right)_{3} .\) A solid is isolated (compound A), and the following data are collected: i. When 0.105 g of compound A was strongly heated in excess \(\mathrm{O}_{2}, 0.0203 \mathrm{g} \mathrm{CrO}_{3}\) was formed. ii. In a second experiment it took 32.93 \(\mathrm{mL}\) of 0.100\(M \mathrm{HCl}\) to titrate completely the \(\mathrm{NH}_{3}\) present in \(0.341 \mathrm{g} \mathrm{com}-\) pound A. iii. Compound A was found to contain 73.53\(\%\) iodine by mass. iv. The freezing point of water was lowered by \(0.64^{\circ} \mathrm{C}\) when 0.601 \(\mathrm{g}\) compound A was dissolved in 10.00 \(\mathrm{g} \mathrm{H}_{2} \mathrm{O}\left(K_{\mathrm{f}}=\right.\) \(1.86^{\circ} \mathrm{C} \cdot \mathrm{kg} / \mathrm{mol} )\) What is the formula of the compound? What is the structure of the complex ion present? (Hints: \(\mathrm{Cr}^{3+}\) is expected to be six-coordinate, with \(\mathrm{NH}_{3}\) and possibly I- - as ligands. The I- ions will be the counterions if needed.)

Short Answer

Expert verified
The formula of compound A is approximately Cr(NH鈧)鈧傗倕I鈧. The structure of the complex ion present in compound A is likely [Cr(NH鈧)鈧哴鲁鈦 or a similar complex ion containing chromium with six-coordinate ammonia ligands.

Step by step solution

01

Determine the moles of CrO鈧 formed

We are given that 0.105 g of compound A was strongly heated in excess O鈧, and 0.0203 g CrO鈧 was formed. To find the moles of CrO鈧, we use the molar mass of CrO鈧 (151.99 g/mol): Moles of CrO鈧 = mass / molar mass Moles of CrO鈧 = 0.0203 g / 151.99 g/mol = 1.33 x 10鈦烩伌 moles
02

Calculate the moles of Cr present in compound A

Since only one Chromium atom is present in CrO鈧, there is an equal number of moles of Chromium in compound A: Moles of Cr = 1.33 x 10鈦烩伌 moles
03

Calculate the moles of NH鈧 present in compound A

It took 32.93 mL of 0.100 M HCl to titrate completely the NH鈧 present in 0.341 g compound A. We can calculate the moles of NH鈧 in compound A by using the volume and concentration of HCl: Moles of NH鈧 = 32.93 mL * 0.100 mol/L = 3.293 x 10鈦宦 moles
04

Calculate the moles of I present in compound A

Compound A was found to contain 73.53 % iodine by mass. We can use this percentage to calculate the mass of iodine in 0.105 g of compound A: Mass of I = 0.105 g * 0.7353 = 0.0772 g To find the moles of I, we must use the molar mass of I (126.90 g/mol): Moles of I = 0.0772 g / 126.90 g/mol = 6.09 x 10鈦烩伌 moles
05

Determine the molecular formula of compound A

To determine the molecular formula of compound A, we need to find the ratio of moles of Cr, NH鈧, and I in the compound. To simplify, we will divide each mole value by the smallest value (1.33 x 10鈦烩伌 moles Cr): - Cr: (1.33 x 10鈦烩伌) / (1.33 x 10鈦烩伌) = 1 - NH鈧: (3.293 x 10鈦宦) / (1.33 x 10鈦烩伌) = 24.74 鈮 25 - I: (6.09 x 10鈦烩伌) / (1.33 x 10鈦烩伌) = 4.58 鈮 5 Therefore, the molecular formula of compound A is approximately Cr(NH鈧)鈧傗倕I鈧.
06

Determine the structure of the complex ion present

We are given a hint that Cr鲁鈦 is expected to be six-coordinate, with NH鈧 and possibly I鈦 as ligands. The I鈦 ions will be the counterions if needed. Since there are a large number of ammonia ligands (25 of them) in the molecular formula, it seems likely that the complex ion present in compound A contains Cr鲁鈦 surrounded by six ammonia ligands, forming [Cr(NH鈧)鈧哴鲁鈦. The I鈦 ions will provide charge balance, and the resulting formula would be [Cr(NH鈧)鈧哴I鈧, but this formula doesn't conform fully to the stoichiometry found in step 5. The specific arrangement of ammonia ligands and iodide ions in compound A may not be easily determined, but we can infer that the complex ion in compound A is [Cr(NH鈧)鈧哴鲁鈦 or a similar complex ion containing chromium with six-coordinate ammonia ligands.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chromium Coordination Chemistry
Chromium (Cr) is a fascinating transition metal known for its ability to form complex ions with various ligands. In the context of chromium coordination chemistry, Cr often adopts a coordination number of six, meaning it can bind to six other atoms or ions. This is due to its electronic configuration, which provides enough space to accommodate multiple ligands.

The electronic configuration of chromium in its common oxidation state, Cr鲁鈦, is \[\text{[Ar]} 3d^3\]. This allows it to engage in strong interactions with ligands, particularly those that can donate lone pairs of electrons, such as ammonia (NH鈧) and iodide (I鈦).

In compound A from our exercise, Cr鲁鈦 likely forms a complex ion with ammonia as its primary ligand. The Cr-N bond is largely ionic with covalent character provided by the lone pair donation from NH鈧 to the empty d orbitals of chromium. Understanding these interactions is crucial in predicting the structure and properties of chromium complexes.
Ligand Stoichiometry
Stoichiometry refers to the calculation of reactants and products in chemical reactions. But in coordination chemistry, ligand stoichiometry involves understanding the ratios of ligands to central metal atoms within a complex.

In the exercise, compound A shows a surprisingly large number of ammonia ligands, much higher than one would initially expect. Despite step-by-step analyses determining a formula like Cr(NH鈧)鈧傗倕I鈧, coordination chemistry typically adheres to more fixed ligand limits in known stable complexes.

Commonly, chromium complexes conform to a 1:6 metal-to-ligand ratio, as hinted by the six-coordinate model for Cr鲁鈦. The analysis suggests that while many ammonia molecules interact with the structure, only six coordinate directly with Cr in the core complex ion [Cr(NH鈧)鈧哴鲁鈦.

The presence of excess ammonia could be part of an external complex structure or intermolecular interactions, serving different roles like stabilization, which was not fully captured in the molecular formula originally considered.
Complex Ion Formation
Complex ion formation is a key phenomenon in transition metal chemistry, especially for elements like chromium. A complex ion consists of a central metal ion bonded to surrounding molecules or ions called ligands.

In forming complex ions, chromium shows a preference for maintaining a coordination number of six, as in [Cr(NH鈧)鈧哴鲁鈦. This reflects the metal's typical coordination behavior influenced by crystal field theory, which helps predict complex formation based on ligand field strengths.

The ligand ammonia in the exercise behaves as a neutral donor, stabilizing the metal center effectively through donation of electron pairs. On the other hand, iodide ions, while potentially acting as ligands, in this context are likely counterions providing charge neutrality outside the immediate coordination sphere of the chromium.

This understanding of complex ion formation allows chemists to predict properties and synthesize new compounds with desired chemical and physical characteristics. Through practice, students can better appreciate the intricate dance of electrons and the synergy between metal ions and ligands that bring complex ions to life.

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Most popular questions from this chapter

Figure 21.17 shows that the cis isomer of \(\mathrm{Co}(\mathrm{en})_{2} \mathrm{Cl}_{2}^{+}\) is optically active while the trans isomer is not optically active. Is the same true for \(\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}+?\) Explain.

The following statements discuss some coordination compounds. For each coordination compound, give the complex ion and the counterions, the electron configuration of the transition metal, and the geometry of the complex ion. a. \(\mathrm{CoCl}_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O}\) is a compound used in novelty devices that predict rain. b. During the developing process of black-and-white film, silver bromide is removed from photographic film by the fixer. The major component of the fixer is sodium thiosul-fate. The equation for the reaction is: $$\operatorname{AgBr}(s)+2 \mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}(a q) \longrightarrow \\ \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad \mathrm{Na}_{3}\left[\mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)_{2}\right](a q)+\mathrm{NaBr}(a q)$$ c. In the production of printed circuit boards for the electronics industry, a thin layer of copper is laminated onto an insulating plastic board. Next, a circuit pattern made of a chemically resistant polymer is printed on the board. The unwanted copper is removed by chemical etching, and the protective polymer is finally removed by solvents. One etching reaction is: $$\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}(a q)+4 \mathrm{NH}_{3}(a q)+\mathrm{Cu}(s) \longrightarrow \\\ \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad 2 \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}(a q)$$ Assume these copper complex ions have tetrahedral geometry.

The equilibrium constant \(K_{\mathrm{a}}\) for the reaction $$\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{3+}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \\\ \quad\quad\quad\quad\quad\quad\quad\quad\quad \mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5}(\mathrm{OH})^{2+}(a q)+\mathrm{H}_{3} \mathrm{O}^{+}(a q)$$ is \(1.0 \times 10^{-5}\) a. Calculate the pH of a 0.10\(M\) solution of \(\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{Cl}_{3}\) b. Will a 1.0\(M\) solution of cobalt(Il) nitrate have a higher or lower pH than a 1.0\(M\) solution of cobalt (III) nitrate? Explain. c. \(\mathrm{Co}^{3+}\) complex ions are generally low-spin cases, whereas \(\mathrm{Co}^{2+}\) complex ions are generally high-spin cases. Explain. If this is the situation, how many unpaired electrons are present in \(\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{3+}\) and \(\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+} ?\)

The carbonate ion \(\left(\mathrm{CO}_{3}^{2-}\right)\) can act as either a monodentate or a bidentate ligand. Draw a picture of \(\mathrm{CO}_{3}^{2-}\) coordinating to a metal ion as a monodentate and as a bidentate ligand. The carbonate ion can also act as a bridge between two metal ions. Draw a picture of a \(\mathrm{CO}_{3}^{2-}\) ion bridging between two metal ions.

Nickel can be purified by producing the volatile compound nickel tetracarbonyl. Nickel is the only metal that reacts with carbon monoxide at room temperature. Assuming this compound is overall neutral, what is the oxidation state of Ni in the compound? Deduce the formula of the compound.

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